proof of ratio test
Assume k<1. By definition ∃N such that
n>N⟹|an+1an-k|<1-k2⟹|an+1an|<1+k2<1
i.e. eventually the series |an| becomes less than a convergent geometric series
, therefore a shifted subsequence
of |an| converges by the comparison test
. Note that a general sequence
bn converges iff a shifted subsequence of bn converges. Therefore, by the absolute convergence theorem, the series an converges.
Similarly for k>1 a shifted subsequence of |an| becomes greater than a geometric series tending to ∞, and so also tends to ∞. Therefore an diverges.
Title | proof of ratio test |
---|---|
Canonical name | ProofOfRatioTest |
Date of creation | 2013-03-22 12:24:46 |
Last modified on | 2013-03-22 12:24:46 |
Owner | vitriol (148) |
Last modified by | vitriol (148) |
Numerical id | 6 |
Author | vitriol (148) |
Entry type | Proof |
Classification | msc 40A05 |
Classification | msc 26A06 |